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Home , Barium nitrate, Silver chromate

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Solution Stoichiometry Worksheet

Solve the following solutions Stoichiometry problems:

1. How many grams of chromate will precipitate when 150. mL of 0.500 M silver are added to 100. mL of 0.400 M ?

2 AgNO3(aq) + K2CrO4(aq)  Ag2CrO4(s) + 2 KNO3(aq)

0.150 L AgNO3 0.500 moles AgNO3 1 moles Ag2CrO4 331.74 g Ag2CrO4 = 12.4 g Ag2CrO4 1 L 2 moles AgNO3 1 moles Ag2CrO4

0.100 L K2CrO4 0.400 moles K2CrO4 1 moles Ag2CrO4 331.74 g Ag2CrO4 = 13.3 g Ag2CrO4 1 L 1 moles K2CrO4 1 moles Ag2CrO4

2. How many mL of 0.280 M nitrate are required to precipitate as barium all the sulfate from 25.0 mL of 0.350 M aluminum sulfate? (93.8 mL )

3 Ba(NO3)2(aq) + Al2(SO4)3(aq)  3 BaSO4(s) + 2 Al(NO3)3(aq)

0.0250 L Al2(SO4)3 0.350 moles Al2(SO4)3 3 moles Ba(NO3)2 1 L = 0.0938 L Ba(NO3)2 1 L 1 moles Al2(SO4)3 0.280 moles Ba(NO3)2

3. 25.0 mL of 0.350 M NaOH are added to 45.0 mL of 0.125 M copper (II) sulfate. How many grams of copper (II) hydroxide will precipitate?

2 NaOH(aq) + CuSO4(aq)  Cu(OH)2(s) + Na2SO4(aq)

0.0250 L NaOH 0.350 moles NaOH 1 moles Cu(OH)2 97.57 g Cu(OH)2 = 0.427 g Cu(OH)2 1 L NaOH 2 moles NaOH 1 mole Cu(OH)2

0.0450 L CuSO4 0.125 moles CuSO4 1 moles Cu(OH)2 97.57 g Cu(OH)2 = 0.549 g Cu(OH)2 1 L NaOH 1 moles CuSO4 1 mole Cu(OH)2

4. What volume of 0.415 M will be required to precipitate as all the bromide in 35.0 mL of 0.128 M calcium bromide?

2 AgNO3(aq) + CaBr2(aq)  Ca(NO3)2(aq) + 2 AgBr(s)

0.0350 L CaBr2 0.128 moles CaBr2 2 moles AgNO3 1 L AgNO3 = 0.0216 L AgNO3 1 L CaBr2 1 moles CaBr2 0.415 mole AgNO3

5. What volume of 0.496 M HCl is required to neutralize 20.0 mL of 0.809 M sodium hydroxide?

HCl(aq) + NaOH(aq)  NaCl(aq) + H(OH)(l)

0.0200 L NaOH 0.809 mole NaOH 1 mole HCl 1 L HCl = 0.0326 L HCl 1 L NaOH 1 mole NaOH 0.496 mole HCl

6. How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid)

2 HCl(aq) + Na2CO3(s)  2 NaCl(aq) + H2CO3(aq)

1.25 g Na2CO3 1 mole Na2CO3 2 mole HCl 1 L HCl = 0.0330 L HCl 105.99 g Na2CO3 1 mole Na2CO3 0.715 mole HCl

7. What minimum number of grams of oxalic acid monohydrate, H2C2O4• H2O, would you specify for a of no fewer than 15.0 mL of 0.100 M NaOH? Both of the hydrogen’s from oxalic acid are replaceable in this reaction.

H2C2O4• H2O(aq) + 2 NaOH(aq)  Na2C2O4• H2O(aq) + 2 H(OH)(l)

0.0150 L NaOH 0.100 mole NaOH 1 mole H2C2O4• H2O 108.06 g H2C2O4• H2O = 0.0810 g H2C2O4• H2O 1 L NaOH 2 mole NaOH 1 mole H2C2O4• H2O

8. How many grams of magnesium hydroxide will precipitate if 25.0 mL of 0.235 M are combined with 30.0 mL of 0.260 M potassium hydroxide?

Mg(NO3)2(aq) + 2 KOH  2 KNO3(aq) + Mg(OH)2(s)

0.0250 L Mg(NO3)2 0.235 mole Mg(NO3)2 1 mole Mg(OH)2 58.33 g Mg(OH)2 = 0.343 Mg(OH)2 1 L Mg(NO3)2 1 mole Mg(NO3)2 1 mole Mg(OH)2

0.0300 L KOH 0.260 mole KOH 1 mole Mg(OH)2 58.33 g Mg(OH)2 = 0.227 g Mg(OH)2 1 L KOH 2 mole KOH 1 mole Mg(OH)2

9. 60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead (II) nitrate. How many grams of lead (II) iodide will precipitate?

2 KI(aq) + Pb(NO3)2(aq)  2 KNO3(aq) + PbI2(s)

0.0600 L KI 0.322 mole KI 1 mole PbI2 461.00 g PbI2 = 4.45 g PbI2 1 L KI 2 mole KI 1 mole PbI2

0.0200 L Pb(NO3)2 0.530 mole Pb(NO3)2 1 mole PbI2 461.00 g PbI2 = 4.89 g PbI2 1 L Pb(NO3)2 1 mole Pb(NO3)2 1 mole PbI2